1.2: Exponents and Radicals

Goal: Understand what $a^{m/n}$ means and how to manipulate such expressions.

Integer Exponents


If $a \in \mathbb{R}$, then the $n$th power of $a$ is \[a^n = a\cdot a \cdot \cdots a\]

Notice how exponents are shorthand for multiplication. This means everything within an exponent is a factor. Keep this in mind.

Evaluate the following:
  1. $\left(\frac{1}{2}\right)^5$
  2. $(-3)^4$
  3. $-3^4$
Zero and negative exponents are defined as \[a^0 = 1 \qquad \qquad a^{-n} = \dfrac{1}{a^n}\]
Evaluate the following:
  1. $\left(\dfrac{100}{101}\right)^0$
  2. $x^{-1}$
  3. $(-2)^{-3}$

Laws of Exponents


In the previous problems, we expanded the exponent into the multiplications. Here are seven shortcuts you can use with exponents.

Laws of Exponents
  1. $a^ma^n = a^{m+n}$
  2. $\dfrac{a^m}{a^n} = a^{m-n}$
  3. $(a^m)^n = a^{mn}$
  4. $(ab)^n = a^nb^n$
  5. $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$
  6. $\left(\dfrac{a}{b}\right)^{-n} = \dfrac{b^n}{a^n}$
  7. $\dfrac{a^{-n}}{b^{-m}} = \dfrac{b^m}{a^n}$

Notice how all of these laws require you to have factors. If you have terms, you cannot use these laws.

Before using Laws of Exponents, get into the habit of asking "Do I have factors? If so, identify the factors."

A student wishes to simplify the expression in the following way: \[(a+b)^2 = a^2 + b^2\] Why can't they do this?
Look within the parenthesis. Since $a$ and $b$ are terms within the local context of the parenthesis, no exponent laws can apply! Therefore, the student misused Laws of Exponents (LoE) #4.
Simplify the following:
  1. $x^4x^7$
  2. $\dfrac{y^4}{y^{-7}}$
  3. $(b^4)^5$
  4. $(3x)^3 \ \ $ (hint: what are the factors?)
  5. $\left(\dfrac{x-1}{2}\right)^5$
  6. $2x^2(x-2)^3(3x^4(x-2))^3$
  7. $\left(\dfrac{x}{y}\right)^3\left(\dfrac{xy^2}{z}\right)^4$
  8. $\dfrac{6t^{-4}(2x+3)}{2t^2(2x+3)^{-2}}$
  9. $\left(\dfrac{y}{3z^3}\right)^{-2}$
  10. $\left(\dfrac{(x-1)^{-1}(x+2)}{(x+2)^2}\right)^2$

Here's solutions with laws listed out whenever an equal sign is used.
  1. $x^4\cdot x^7 = x^{4 + 7} = x^{11}$ by LoE 1.
  2. \begin{align} \dfrac{y^4}{y^{-7}} &= y^{4 - (-7)} && \text{LoE 2} \\&= y^{4 + 7} && \text{Negative Law 2} \\&= y^{11} \end{align}
  3. \begin{align} (b^4)^5 &= b^{4 \cdot 5} && \text{LoE 3} \\&= b^{20} \end{align}
  4. The factors within the parenthesis are $3$ and $x$. Use LoE 4 \[(3x)^3 = 3^3 \cdot x^3 = 27x^3\]
  5. The numerator is comprised of terms. If you read LoE 5 in English, it says the global context of the numerator must receive the power. Therefore group the terms together into one factor $(x-1)$: \[ \left(\dfrac{x - 1}{2}\right)^2 = \dfrac{(x-1)^5}{2^5} = \dfrac{(x-1)^5}{32} \]
  6. This problem looks more difficult than the others. But remember: look for the factors! They break a problem into simpler ones you just saw. This strategy is called divide and conquer in computer science. Within the rightmost factor, look within for factors as well. We see the factors are $3, x^4$ and $(x-2)$. \begin{align} 2x^2\cdot(x-2)^3 \cdot(3x^4(x-2))^3 &= 2x^2\cdot(x-2)^3\cdot 3^3 \cdot (x^4)^3 \cdot (x-2)^3 && \text{LoE 4} \\&= 2\cdot 27 \cdot x^{2}\cdot x^{12} \cdot (x-2)^{3 + 3} && \text{LoE 1, 3 and comm. prop.} \\&= 54 x^{2 + 12}(x-2)^6 && \text{LoE 1} \\&= 54x^{14}(x-2)^6 \end{align}
  7. Again, divide and conquer by looking at the factors (in global ctx.). Each fraction in parenthesis is a factor. Focus within each. Remember, LoE 5 says the global ctx. of the numerator and denominator needs to receive the power. \begin{align} \left(\dfrac{x}{y}\right)^3\left(\dfrac{xy^2}{z}\right)^4 &= \dfrac{x^3}{y^3} \cdot \dfrac{(xy^2)^4}{z^4} && \text{LoE 5} \\&= \dfrac{x^3}{y^3} \cdot \dfrac{x^4 (y^2)^4}{z^4} && \text{LoE 4} \\&= \dfrac{x^3}{y^3} \cdot \dfrac{x^4 y^8}{z^4} && \text{LoE 3} \\&= \dfrac{x^3x^4y^8}{y^3z^4} && \text{Fraction Law 1} \\&= \dfrac{x^7y^5}{z^4} && \text{LoE 1 + 2} \end{align}
  8. Let's deal with the negative exponents first. \begin{align} \dfrac{6t^{-4}(2x+3)}{2t^2(2x+3)^{-2}} &= \dfrac{6(2x+3)(2x+3)^2}{2t^4t^2} && \text{LoE 7} \\&= \dfrac{6(2x+3)^{1 + 2}}{2t^{4 + 2}} && \text{LoE 1} \\&= \dfrac{3(2x+3)^3}{t^6} && \text{Fraction Law 5 (cancellation law)} \end{align}
  9. The more you get into the habit of identifying factors, the more you will realize each exponent simplifcation problem is really just two steps: factor identification then matching the factor form with the correct Law of Exponents. \begin{align} \left(\dfrac{y}{3z^3}\right)^{-2} &= \dfrac{(3z^3)^2}{y^2} && \text{LoE 6} \\&= \dfrac{3^2(z^3)^2}{y^2} && \text{LoE 4} \\&= \dfrac{9z^6}{y^2} && \text{LoE 3} \end{align}
  10. Refer to in-class notes.

Radicals


What does $2^{\frac{4}{5}}$ mean? Well, $2^{\frac{4}{5}} = 2^{4 \cdot \frac{1}{5}}$. Laws of Exponents takes care of the 4.

We just need to give meaning to the exponent $\frac{1}{5}$ or in general, $\frac{1}{n}$ where $n \in \mathbb{N}$.

Let's do so.

Recall that \[\sqrt{a} = b \qquad \text{means} \qquad b^2 = a, \ b \geq 0\]

What is $\sqrt{4}$ and why?
Changing the $2$ into any $n \in \mathbb{N}$:
If $n \in \mathbb{N}$ then the principal nth root of $a$ is defined as \[\sqrt[n]{a} = b \qquad \text{means} \qquad b^n = a\] If $n$ is even, we must have $a \geq 0, b \geq 0$.
Simplify:

Rational Exponents


For any rational exponent $m/n$ in lowest terms, we define \[a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}\]

Denominator becomes the principal nth root, numerator becomes the exponent within the root.

Laws of Exponents also hold for rational exponents.

Simplify:
  1. $4^{1/2}$
  2. $8^{2/3}$
  3. $125^{-1/3}$
  4. $(2x^2 + 2)^{1/3}(2x^2 + 2)^{7/3}$
  5. $\dfrac{(x-2)^{2/5}(2x-1)^{7/5}}{(x-2)^{3/5}}$

    Each equal sign is either a law or a definition.
  1. $4^{1/2} = \sqrt[2]{4^1} = 2$
  2. $8^{2/3} = \sqrt[3]{8^2} = \sqrt[3]{64} = \sqrt[3]{2^6} = 2^{\frac{6}{3}} = 2^2 = 4$
  3. \begin{align} 125^{-1/3} &= \dfrac{1}{125^{\frac{1}{3}}} && \text{Definition of negative exponent} \\&= \dfrac{1}{\sqrt[3]{125}} && \text{Definition of rational exponent} \\&= \dfrac{1}{\sqrt[3]{5^3}} && \text{Prime Factorization} \\&= \dfrac{1}{5} \end{align}
  4. Remember, if you don't know what to do, look for the factors! Clearly, LoE 1 should be applied. \begin{align} (2x^2 + 2)^{1/3}(2x^2 + 2)^{7/3} &= (2x^2 + 2)^{\frac{1}{3} + \frac{7}{3}} && \text{LoE 1} \\&= (2x^2 + 2)^{\frac{1 + 7}{3}} && \text{Fraction Law 3} \\&= (2x^2 + 2)^{\frac{8}{3}} \end{align}
  5. Refer to in-class notes.